Ponder This Challenge - March 2003 - Fencing a pasture

Ponder This Challenge:

This month's puzzle was supplied by Vladimir Sedach.

We have 100 meters of fencing, which we use to enclose a pasture. Our fence must begin and end at the oak tree. Ground south of the oak tree, or less than 20 meters north of the oak tree, is worth $100 per square meter; but ground more than 20 meters north of the oak tree is worth$200 per square meter.

What shape maximizes the total value of the enclosed pasture, and what is this value?

Clarifications:
The "oak tree" is a single point. "20 meters north" refers to an east-west line passing 20 meters north of the tree. "Begin and end at the oak tree" means that both ends of the fence touch this point. All of this takes place on a flat Earth.

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Solution

• Click here to view the solution

  We use physical approach: maximum area inside the rope is when the rope is
  stable, and the pressure (price) of the air in upper area (y > 20) is 2 times more
  than in lower (y < 20).

  Radius of the stable rope is in reverse proportion to the air pressure, and
  is constant if the pressure is constant along the rope. So we have one
  circle with radius r in the area (y > 20) and two circles with radius r2 in
  the area (y < 20). Circles r and r2 have the same tangent at y = 20. See the
  picture for clarity.

  r2 * sin(b) = (r2 - r) * cos(a);

  r2 * cos(b) = yp + r2 * sin(a);

  r * sin(a) = yc - yp;

  r * (pi + 2*a) + r2 * (pi - 2*a - 2*b) = len;

  r2 = r * k;

  where:

  yp = 20

  yc - center of circle r

  r - radius of the circle lying above y=yp

  r2 - radius of two circles lying below y=yp, and crossing at (0, 0)

  a - angle between axis x and radius r2 to the point of contact

  b - angle between axis y and radius r2 to the point (0, 0)

  len = 100 - length of the rope

  k = 2 - p(y > 20) / p(y < 20)

  Solution: r=14.23185486, yc=22.38258653

  Area above the line: ahi = (1/2)*r*r*(pi + 2*a) + r*cos(a)*r*sin(a) = 385.6571522

  Area below the line: alo = r2*r2*((pi/2)-b-a) - r*cos(a)*r*sin(a) - (r2-r)*cos(a)*( r2*cos(b) - (r2-r)*sin(a) ) = 371.2511721

  price = 200 * ahi + 100 * alo = 114256.5477

  ---

  If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com

Solvers

• Stephane Peysson (03.04.2003@12:46:39PM EST)
• Michael Brand (03.04.2003@01:41:00PM EST)
• Michael Swart (03.04.2003@04:27:17PM EST)
• Jayavel Sounderpandian (03.06.2003@06:01:01PM EST)
• Hagen von Eitzen (03.08.2003@11:24:34AM EST)
• Steve Horvath (03.10.2003@09:05:29AM EST)
• John G. Fletcher (03.10.2003@03:55:13PM EST)
• Chris Shannon (03.10.2003@05:26:00PM EST)
• Alex Proudfoot (03.13.2003@06:53:42PM EST)
• Aditya Utturwar (03.17.2003@11:21:19AM EST)
• Rainer Rosenthal (03.17.2003@05:42:29PM EST)
• Morgan Sherman (03.19.2003@08:07:24PM EST)
• Sebastian Casalaina-Martin (03.19.2003@08:07:24PM EST)
• Maxim G. Smith (03.20.2003@02:59:47PM EST)
• Rocke Verser (03.21.2003@05:59:27AM EST)
• Andreas Knappe (03.21.2003@07:39:39AM EST)
• Bob Stewart (03.21.2003@11:50:09AM EST)
• Florbela Duarte (03.21.2003@06:51:06PM EST)
• Gyozo Nagy (03.23.2003@04:19:25PM EST)
• Donald T. Dodson (03.25.2003@09:44:10AM EST)
• Nachiket Bapat (03.28.2003@06:06:30PM EST)
• Amos Guler (03.30.2003@02:29:21AM EST)
• Troels Nørgaard (03.31.2003@06:21:42AM EST)
• Nis Jorgensen (03.31.2003@11:42:39AM EST)